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Ok. When I say "infinity," I mean an infinitely large number (such as $9999999\ldots$)

So we all know the common proof for $0.999\ldots = 1$.

If not, here it is:

$$ \begin{align*} x &= 0.999\ldots \\ 10x &= 9.999\ldots \\ 10x-x &= 9x \\ 9.999\ldots - 0.999\ldots &= 9 \end{align*} $$

Therefore, $9x = 9, x = 1$.

Using this logic, it can be proved that $0.000\ldots1 = 0$ because

$$ \begin{align*} 1 - 1 &= 0 \\ 1 - 0.999\ldots &= 0.000\ldots1 \end{align*} $$

and since $1 = 0.999\ldots$, that means that $0 = 0.000\ldots1$.

Now, if we were to take any number and divide it by an infinitely large number, then the answer would eventually consist of $0.000000\ldots1$ (or at least have an infinite series of zeros before a number other than one). Since $0.000\ldots1 = 0$, this must mean that $\frac{1}{\infty} = 0$.

Is this correct?

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    $\begingroup$ There is no such thing as $0.0\ldots 01$ $\endgroup$ – Hagen von Eitzen May 28 '17 at 2:41
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    $\begingroup$ no such thing as $9999999\ldots$ either. If you want to use it, you need to define it in a meaningful way. Same applies to "infinitely large number". $\endgroup$ – Mirko May 28 '17 at 4:22
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    $\begingroup$ There's no such thing as an infinitely large number. There is no such thing as 999999...... there's no such thing as 0.00000......1. And there is no such thing as 1/infinity. No. That is not correct. $\endgroup$ – fleablood May 28 '17 at 5:10
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    $\begingroup$ To slightly correct Hagen and Mirko, there are no real numbers with such decimal representation. Slightly related to your question is my answer and the comments below it here: math.stackexchange.com/questions/979177/… $\endgroup$ – Asaf Karagila May 28 '17 at 5:13
  • $\begingroup$ @Asaf: To be fair, by the usual definition of numerals (specifically, a $\mathbb{Z}$-indexed sequence of digits, with the point being between indexes $0$ and $-1$), neither notation is shorthand for a decimal numeral either. $\ldots 9999$ is, though. $\endgroup$ – Hurkyl May 29 '17 at 6:59

10 Answers 10

15
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(Even though the mathematics is incorrect, +1 for a well-asked question!)

Your core misunderstanding seems to be the definition of $0.999\dots$. By definition, that number equals $\frac9{10} + \frac9{10^2} + \frac9{10^3} + \cdots$, an infinite series (which equals $1$, by the proof you gave). But my point is that it has a specific mathematical definition; it's not just "notation for an idea".

On the other hand, you seem to be thinking of $0.000\dots01$ as an infinite string of $0$s followed by a $1$. Such an object does not have any mathematical meaning, and so reasoning with it will not be mathematically valid. Alternatively, you might be thinking of that expression as "what happens when I take $0.000\dots01$, with a finite number (call it $n$) of $0$s, and let $n$ get larger and larger". While I wouldn't use that notation for this idea, the idea itself is perfectly valid and leads to the true mathematical fact $$ \lim_{n\to\infty} \frac1{10^n} = 0. $$ In either case, your equation $1 - 0.999\dots = 0.000\dots1$ is not correct, because the number $0.999\dots$ does not have any last $9$ that would yield a $1$ after the subtraction.

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All you can say is

$$\lim_{x\to x_0}f (x)=\infty\implies \lim_{x\to x_0}\frac {1}{f (x)}=0$$ or

$$\lim_{n\to+\infty}u_n=\infty\implies \lim_{n\to+\infty}\frac {1}{u_n}=0.$$

If a number becomes greater and greater, its reciprocal becomes smaller and smaller (obvious).

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  • $\begingroup$ Thanks for reciprocal. +1. $\endgroup$ – hamam_Abdallah May 28 '17 at 2:47
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Do you consider that $9999999\ldots = 99999999\ldots$ ? Indeed they seem to be the same number, an infinite decimal of repeating $9$'s. Hence their difference $99999999\ldots - 9999999\ldots = 0$, and:

\begin{align*} x &= 9999999\ldots \\ 10x &= 99999999\ldots \\ 10x-x &= 9x \\ 99999999\ldots - 9999999\ldots &= 0 \end{align*}

Therefore $9x=0$, and $x=0$. This proves that $9999999\ldots =0$.

Bonus: Since $\frac1\infty=0$, and $9999999\ldots=\infty$ we also get that $0=\frac1\infty=\frac1{9999999\ldots}=\frac10$.

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Infinity is not a number, so it doesn't work in an algebraic setting. However it can certainly be treated as a limit, and

$$\lim_{x \to \infty} \frac1x = 0$$

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The answers above use limits to deal with infinities. This is the rigorous approach to the Calculus that was formulated in the 19th Century. What you are trying to accomplish in the opening post reminds me of Abraham Robinson's non-standard analysis, developed in the mid 20th Century. In this approach, the reciprocal of an infinitely large number is an infinitesimal -- a number infinitely close -- but not equal -- to zero.

A very readable introduction to non-standard analysis can be found here. Judging by your approach to the infinite, I suspect you may find the information in the opening chapters very exciting.

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0
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You can't divide by infinity as it is not a number. It just doesn't work with any logic, even that proof.

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0
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In this case for the sequence $n\to\infty$ then $\frac 1n\to 0$.

Or for continuous variable $\lim\limits_{x\to+\infty}\frac 1x=0$

So in a sense $\frac 1{\infty}=0$ but we are not manipulating numbers anymore, just the concept of limit.

Note that stating the reverse is more delicate, since we use to give a sign to infinity.

Both $\lim\limits_{x\to+\infty} \frac 1x=\lim\limits_{x\to-\infty}\frac 1x=0$ but we cannot conclude $\frac 10=\infty$ because theoretically (at least for the usual real numbers) we would have to separate the positive case and the negative case.

$\lim\limits_{x\to 0^+}\frac 1x=+\infty$ and $\lim\limits_{x\to 0^-}\frac 1x=-\infty$

What's important to keep in mind, is that when we are casually talking about $\frac 10$ or $\frac 1\infty$, we are in reality talking about limits in a well defined theoretic context. So there is no issue if we respect these fundamentals, but there could be if we go too casual and start enjoying calculating wildly with these non-numbers.

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We usually define an operation for a set. Suppose the set that we wish to discuss is the set of real number, $\mathbb{R}$. Then, a possible short answer is since $\infty$ is not a real number, hence, it does not make sense to talk about such thing.

what we can say is

$$\lim_{x \rightarrow \infty} \frac1x = 0$$

While $\infty$ is not a real number, there is a generalization called extended real number which is useful in describing various limiting behaviors in calculus and mathematical analysis, especially in the theory of measure and integration. It include $\infty$ in the set of discussion.

It is commonly denoted by $\bar{\mathbb{R}}$ or $[-\infty, \infty]$.

We do define $$\frac{1}{\infty}=0$$ for extended real number. However, be careful that normal operations that you expect might not hold, in particular,

$$\frac10 \neq \infty.$$

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The mathematics is correct except for an ambiguity in your notation. Notice that when you write a number (an infinitesimal) of the form $0.000\ldots 01$ with an infinite number of digits, there is a last zero before a final $1$. On the other hand when you write a number of the form $999\ldots$ there is an ambiguity with your notation that should be resolved by specifying a final digit $9$, as in $999\ldots9$ (again with infinite number of digits). Then $999\ldots9$ is infinite while $0.000\ldots01$ is a nonzero infinitesimal, and their reciprocals are respectively a nonzero infinitesimal and an infinite number. For additional details see this article by Lightstone. The correct relationship between 0.999...9 (with an infinite number of 9s) and 1 is not that of equality but rather $$\mathbb{st}(0.999...9)=1$$ where $\mathbb st$ is the standard part. For the same reason, if $\infty$ denotes an infinite number, the reciprocal $$\frac{1}{10^\infty}$$ is not zero but rather infinitesimal (only its standard part is zero).

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We don't know, and it's depending on quite a lot of factors, too. Like, what infinity are you using? Bear in mind that $\lim\limits_{n\to \infty}n$ results into $\infty$ but $\lim\limits_{n \to \infty}n^2$ does not equal the vanilla $n$-variant, and both get dirrefent answers. Also, you cannot get rational results using infinities. We can get close, though, using limits: $$\lim\limits_{x \to \infty} \frac{1}{x} = 0$$ That basicly means that if we get $x$ reeaally big, $\frac{1}{x}$ will approach 0. We can say it won't actually reach 0 though, not even when $x = \infty$. Otherwise we'd get $\frac{1}{\infty} = 0$ resulting in both $1 = \infty * 0$ and $\frac{1}{0} = \infty$, which both are incorrect statements (the first one because $a = a * 1 = a * 0 * \infty = 0 * \infty = 1$, which is incorrect, and the second one because we don't divide by $0$). Furthermore, the "mistake" made by the proof that $0.999... = 1$ is that they don't use limits (for further explanation on that, look at the question I asked earlier on my profile, lots of answers there), and, you should keep in mind that you don't know if $\infty = 10^x$, so $\frac{1}{\infty}$ does not necessarily have to be $0.000...01$. That should answer most of your questions.

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  • $\begingroup$ -1 for your first three sentences. $\endgroup$ – Alex S Nov 16 '17 at 2:18
  • $\begingroup$ I just changed it ;) better now? @AlexS $\endgroup$ – SuperSjoerdie Nov 16 '17 at 5:40
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    $\begingroup$ Sorry, still not on board. $\lim_{n\to\infty} n$ is the same infinity as $\lim_{n\to\infty} n^2$. I see that what you are saying is that doing arithmetic with infinity is not really useful, but we have to be precise here. You should also note that $0.000...01$ is not a number. $\endgroup$ – Alex S Nov 16 '17 at 13:11

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